Yesterday in class this question came up by one of the students in the class. Does anyone know why there is a minus sign in front of the W2 in the figure?
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1$\begingroup$ And the TA said? $\endgroup$– Solar MikeCommented Apr 4, 2018 at 15:00
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1 Answer
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If you work out the transfer functions from $d$ to $z_1$ and $z_2$ it can be shown that you get
$$ z_1 = W_1 \frac{1}{1+K\,G} d, $$
$$ z_2 = -W_2 \frac{-K}{1+K\,G} d. $$
The part of the transfer function of $z_1$ without $W_1$ is often called the sensitivity function. For $H_\infty$ you often choose $W_1$ to be a semi-inverse of a desired sensitivity function.
The part of the transfer function of $z_2$, after cancelling the two minuses, without $W_2$ is often called the control sensitivity function (however here they also call it the noise sensitivity function). For $H_\infty$ you again often choose $W_2$ to be semi-inverse of a desired control sensitivity function.
So in order for $W_2$ to have a more meaningful meaning they added a minus sign into its definition, such that the control sensitivity function is weighted by $W_2$ instead of $-W_2$.
PS: by semi-inverse I am roughly referring to a transfer function which is causal/proper, stable (all poles have a nonzero negative real part) and is an approximation of the inverse of a given transfer function.
